Question: Estimating $\ln(0.75)$ using a Taylor polynomial about $x=1$, what is the least degree of the polynomial that assures an error smaller than $0.001$ ? (The $n^{\text{th}}$ derivative of $\ln(x)$ is $(-1)^{n-1}\dfrac{(n-1)!}{x^n}$ for $n\geq1$ ) Choose 1 answer: Choose 1 answer: (Choice A) A $3$ (Choice B) B $4$ (Choice C) C $5$ (Choice D) D $6$
Solution: We will use the Lagrange error bound. Let's assume the polynomial's degree is $n$. The $(n+1)^{\text{th}}$ derivative of $\ln(x)$ is $(-1)^n\dfrac{n!}{x^{n+1}}$. The Lagrange bound for the error assures that $\begin{aligned} |R_n(0.75)|&\leq\left|\dfrac{(-1)^n\dfrac{n!}{z^{n+1}}}{(n+1)!}(0.75-1)^{n+1}\right| \\\\ &=\dfrac{n!\cdot 0.25^{n+1}}{(n+1)!z^{n+1}} \\\\ &=\dfrac{0.25^{n+1}}{(n+1)z^{n+1}} \end{aligned}$ where $0.75\leq z\leq 1$. On this interval, $\dfrac{0.25^{n+1}}{(n+1)z^{n+1}}\leq\dfrac{0.25^{n+1}}{(n+1)0.75^{n+1}}$. Solving $\dfrac{0.25^{n+1}}{(n+1)0.75^{n+1}}<0.001$ using trial and error, we find that $n\geq4$. In conclusion, the least degree of the polynomial that assures our error bound is $4$.